Note. In the above EXAMPLE you might begin at the 4th Course to add the Westings and subtract the Eastings; or at the 6th Course to add the Northings and subtract the Southings; or at the 11th Course to add the Southings and subtract the Northings. So in every Survey some place may be found where you may begin to add and subtract, without running out before you get through all the Courses.

When a Field is very irregularly shaped, it will often happen that parts of the same Area will be contained in several different Products in the Columns of Areas; but in the final result, one Column being subtracted from the other will leave what is included within the boundary Lines of the Field.

DEMONSTRATION. See PLATE III. Fig. 64. and EXAMPLE II.

The Area standing against the 9th Course, which is where the Calculation begins, is the Triangle I2K, all without the Field.

The Area against the 10th Course is the Trapezoid 2KL3, also without the Field.

The Area against the 11th Course is the Trapezoid 4ML3. This is a South Area, and contains a part of the Field and also part of the preceding North Area.

The Area against the 12th Course is the Trapezoid 5NM4, part within and part without the Field.

The Area against the 13th Course is the Trapezoid 6AN5, part within and part without the Field.

The Area against the 1st Course is the Trapezoid 6AB7, part within and part without the Field. This is a North Area and to be ultimately subtracted from the South Areas; but this includes a part of the preceding South Area, viz. the space nAso; it will however be seen hereafter that this same space is included in another South Area. This North Area contains also a part of the first North Area, viz. the space 6n07; but the same space is also included in another South Area.

The Area against the 2d Courses is also a North Area, and is the Trapezoid 7BC8.. This Trapezoid contains the space sBCx, without the Field; the space osxw, within the Field; and the space 7ow8, without the Field. But the space osxw will be contained in the next South Area; and the space 70w8, which was contained in the two first North Areas, will be contain. ed in the next South Area.

By examining the whole Figure, in this manner, it will be seen that the North Areas contain all without the Field that is taken into the Calculation, and some of it twice over; they also contain part of the Area within the Field. The South Areas contain all within the Field, and all without the Field that is contained in the North Areas. They also contain, twice over, so much of the Field as is included in any of the North Areas; and likewise, twice over, that part without the Field which is contained twice in the North Areas. So that subtracting the North from the South Areas leaves double the Area of the Field.

This method of calculating the Area of a Field by the Northings, Southings, Eastings and Westings, divides the Field, with a certain quantity of the adjoining ground, into Right Angled Triangles, Right Angled Trapezoids, Parallelograms, or Squares, as may be seen by the Figures. It may therefore with propriety be called RECTANGULAR SURVEYING.

A USEFUL PROBLEM.

To find the true Area of a Field which has been measured by a Chain too long or too short.

Calculate the Area as if the Chain was of a true length, then institute the following Proportion: As the Square of the length of the true Chain; Is to the Area, as found by the Chain made use of; So is the Square of the length of that Chain; To the true Area of the Field.

EXAMPLE.

Suppose a Field, measured by a Two Rod Chin 3 Inches too long, is found to contain 41 Acres 1 Rood and 33 Rods, what is the true Area?

As the Square of 33 Feet, the true length of a Two Rod Chain; Is to 41 Acres 1 Rood and 33 Rods; So is the Square of 33 Feet 3 Inches, the length of the Chain used in the Survey; To 42 Acres and 13 Rods. 33 Feet 396 Inches. 396x396=156816 Square Inches.

41 Acres 1 Rood 33 Rods-6633 Rods."

33 Feet 3 Inches=399 Inches. 399x399=159201 Square Inches.

159201x6633-156316-6733 Rods. 6733-160=42 Acres 13 Rods, the true Area.

*

PART II.

LAYING OUT LAND.

PROBLEM I. To lay out any number of Acres in the form of a Square.

Annex 5 Cyphers to the number of Acres, which will turn them into Square Links, the Square Root of which will be the Side of the Square in Links.

EXAMPLE. It is required to lay out 810 Acres in the form of a Square.

Answer. Each Side of the Square must be 9000 Links, or 90 Chains.

PROBLEM II. To lay out any number of Acres in the form of a Parallelogram, whereof one Side is given. Divide the number of Acres, when turned into Square Links, by the given Side; the Quotient will be the Side required.

EXAMPLE. What must be the longest side of a Parallelogram, which is to contain 25 Acres, when the shortest side is 5 Chains and 50 Links?

Answer. 2500000-550-4545 Links for the longest Side.

PROBLEM III. To lay out any number of Acres in a Field, 3, 4, 5, 6, &c. times as long as it is broad.

Divide the Acres, when turned into Square Links, by the proportion between the length and breadth; the Square Root of the Quotient will be the shortest Side. EXAMPLE. It is required to lay out 100 Acres 5 times as long as it is broad.

Answer. 10000000÷5=2000000 the Square Root of which is 1414 Links for the shortest Side, and the longest will be 7070 Links.

PROBLEM IV. To make a Triangle which shall contain a given number of Acres, being confined to a cer

tain Base.

Double the given number of Acres, to which annex 5 Cyphers, and divide by the Base; the Quotient will be the Perpendicular in Links.

EXAMPLE. Upon a Base of 40 Chains to lay out 100 Acres in a Triangular from.

Answer. 5000 Links or 50 Chains will be the length of the Perpendicular.

The Perpendicular may be erected from any part of the Base: Thus, the Triangle ABC. See PLATE II. Fig. 55. is the same as ABE, each containg 100 Acres.

When the given Base is so situated that a Perpendicular of sufficient length cannot be erected therefrom, continue the Base as from B to D. Fig. 56. from which erect the Perpendicular DC, and complete the Triangle

PART III.

DIVIDING LAND.

As different Fields are so variously, and many of them irregularly shaped, and as they are required to be divided in many different proportions, it is difficult to give Rules which will apply to particular cases. The business of dividing Land must therefore be left, in a great measure, to the skill and judgment of the Surveyor; who, if he is well acquainted with Trigonometry, and with measuring Land, will not find it difficult after a little practice, to divide a Field in such a manner as shall be desired. If he has before him a Plot of the Field, and knows the number of parts into which it is to be divided, and the proportion which each part is to bear to the others, he will readily find out where the dividing Lines are to be drawn.

A few RULES and EXAMPLES will be given for the general instruction of the Learner.

PROBEM I. To cut off any number of Acres from a Square or Parallelogram.

Say, As the whole number of Acres in the Field; Is to the length of the Square or length or breadth of the Parallelogram; So is the number of Acres proposed to be cut off; To their proportion of the length or breadth.

PROBLEM II. To cut off any number of Acres by a Line proceeding from any Angle of a Triangle.

Measure the Base, or Side opposite the Angle from which the dividing Line is to be drawn; Then say, As the number of Acres in the whole Triangle; Is to the whole Base; So is the given number of Acres; To their part of the Base.

EXAMPLE. See PLATE II. Fig. 57.

In the Triangle ABC, which contains 48 Acres, it is required to cut off 18 Acres, by a Line proceeding from C to the Base AB, which is 40 Chains.

As 48: 40:: 18: 15

Lay 45 Chains on the Base from B to D, and draw the Line CD. The Triangle will then be divided as was proposed; BCD containing 18 Acres.